Question

Application of the First Order ODE

Electric Circuits - RC

Answer 1

http://prntscr.com/m4y4pc

Answer 2

\(L\frac{di}{dt}+Ri=E(t)\)

\(0.1\frac{di}{dt}+(15)i=30\)

\(\frac{di}{dt}+150i=300\)

\(p(t)=150~~~,~~~q(t)=300\)

\(u(t)=e^{\int_{}{}150dt}\)

\(u(t)=e^{150t}\)

\(\frac{d}{dt}(e^{150t}i)=e^{150t}(300)\)

\(i=\frac{1}{e^{150t}}(\int_{}{}e^{150t}300dt)\)

\(i=\frac{1}{e^{150t}}(2e^{150t}+c)\)

\(i=2+\frac{C}{e^{150t}}\)

when \(i=0,t=0\)
\(0=2+c\)

\(c=-2\)

\(i=2-\frac{2}{e^{150t}}\)

when \(t~\rightarrow~\infty\)
\(i=2-\frac{2}{\infty}\)

\(i=2\)

Answer 3

Is my working correct?

Answer 4

http://prntscr.com/m4zdti

Answer 5

\(L\frac{di}{dt}+Ri=E(t)\)

\(20\frac{di}{dt}+2i=120\)

\(\frac{di}{dt}+\frac{1}{10}i=6\)

\(p(t)=\frac{1}{10}~~~,~~~q(t)=6\)

\(u(t)=e^{\int_{}{}\frac{1}{10}dt}\)

\(u(t)=e^{\frac{1}{10}t}\)

\(\frac{d}{dt}(e^{\frac{1}{10}t}i)=e^{\frac{1}{10}t}(6)\)

\(i=\frac{1}{e^{\frac{1}{10}t}}(\int_{}{}e^{\frac{1}{10}t}(6)dt)\)

\(i=\frac{1}{e^{\frac{1}{10}t}}(60e^{\frac{t}{10}}+c)\)

\(i=60+\frac{C}{e^{\frac{t}{10}}}\)

when \(i=0,t=0\)
\(0=60+C\)
\(C=-60\)

\(i=60-\frac{60}{e^{\frac{t}{10}}}\)

Answer 6

Is this one correct too?

Answer 7

@nuts is our resident electrical engineering major

Answer 8

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Shadow
@nuts is our resident electrical engineering major
\(\color{#0cbb34}{\text{End of Quote}}\)
>.< he might be busy at this hour

Answer 9

Well he is your best shot so might as well tag him (:

Answer 10

seems legit enough

Answer 11

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Shadow
Well he is your best shot so might as well tag him (:
\(\color{#0cbb34}{\text{End of Quote}}\)
haha,alright XD

Answer 12

t->infinity current is correct

Answer 13

i can look in more detail if you want... might be too lazy to actually go through the math so i'll just run on simulator lol

Answer 14

just wanna make sure whether the answer was right
hmm,what kind of simulator do u use? :o

Answer 15

in this case, I can't let
\(C=\frac{C}{e^{150t}}\) right?

Answer 16

nvm i didnt' realize you posted 2 problems.

the first one looks ok... the second one is basically a repeat of the first so it's not worth checking lol

Answer 17

\(\color{#0cbb34}{\text{Originally Posted by}}\) @nuts
nvm i didnt' realize you posted 2 problems.

the first one looks ok... the second one is basically a repeat of the first so it's not worth checking lol
\(\color{#0cbb34}{\text{End of Quote}}\)
yeah,if first one was right,
Im pretty the second should be right XD

Answer 18

honestly the closed form solution of RL, RC, and RLC circuits are readily available online... I usually just look them up on electronicsws or on wikipedia

Answer 19

pretty sure**

Answer 20

of course in an ODE class it's more important to derive time domain solutions from first principles (KVL, KCL, and doing the math), but in practice we typically take things to the frequency domain or the s-domain (fourier/laplace transforms respectively) and quickly find the solution

Answer 21

but yeah your answers seem to match reasonable numbers

Answer 22

\(\color{#0cbb34}{\text{Originally Posted by}}\) @nuts
of course in an ODE class it's more important to derive time domain solutions from first principles (KVL, KCL, and doing the math), but in practice we typically take things to the frequency domain or the s-domain (fourier/laplace transforms respectively) and quickly find the solution
\(\color{#0cbb34}{\text{End of Quote}}\)
Thanks for the tip

Answer 23

later you might learn laplace transforms, and you will see how they make solving linear odes a trivial task...

Answer 24

\(\color{#0cbb34}{\text{Originally Posted by}}\) @nuts
but yeah your answers seem to match reasonable numbers
\(\color{#0cbb34}{\text{End of Quote}}\)
yay! Thanks bro!
LOL,just finish chapter 1 >.<
Omg,still got 4 more chapters left.

Answer 25

\(\color{#0cbb34}{\text{Originally Posted by}}\) @nuts
later you might learn laplace transforms, and you will see how they make solving linear odes a trivial task...
\(\color{#0cbb34}{\text{End of Quote}}\)
owh I c,that's interesting :D

Answer 26

yeah basically if you think about it the solution to any ODE comes in the form
\(\Huge e^{(\sigma+i\omega)x}\)
(either a real exponential or a complex exponential which is a real exponential multiplied by a sinusoid).
Recognizing that the derivative of this expression and the integral of this expression are just constant factors, the laplace transform takes differential equations into an s-domain world where everything is thought of in terms of these exponentials making it easier to perform calculus on differential equations (turns calculus into algebra). You then take the inverse laplace transform to get your answer.

Of course you have to worry about convergence and the inverse laplace transform can often be a pain, but with laplace transform tables it's often times faster.

Answer 27

**linear ODE

Answer 28

woah,can't wait to read that chapter,hehe

Thanks for the info

also,thanks for the help ^

Answer 29

yup np